Lecture Notes on Chemistry - General Chemistry, Slides | CHEM 142, Study notes of Chemistry

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Welcome to CHEMISTRY !!! ¥ An Observational Science ¥ An Experimental Science ¥ A Laboratory Science ¥ An Interesting Science ¥ An Important Science ¥ A Hard Science CHEMISTRY The Study of Matter and its Properties, the Changes that Matter Undergoes, and the Energy Associated with those Changes Chemistry Oceanography Atmospheric Sciences Economics Physics Medicine Governments Geology Anthropology Biology Astronomy Politics People Chemistry as the Central Science Chemistry 142 Text: Chemistry - The Molecular Nature of Matter and Change - By Martin Silberberg Chapter #1 : Keys to the Study of Chemistry Chapter #2 : The Components of Matter Chapter #24 : Nuclear Reactions and Their Applications Chapter #3 : Stoichiometry: Mole - Mass Relationships in Chemical Systems Chapter #4 : The Major Classes of Chemical Reactions Chapter #5 : Gases and the Kinetic - Molecular Theory Chapter #6 : Thermochemistry : Energy Flow and Chemical Change Chapter #7 : Quantum Theory and Atomic Structure Chapter #8 : Electron Configuration and Chemical Periodicity Chemistry Homework !!! Linus Pauling - 1967 Chemistry is not a spectator sport , you must become involved, and that means that you must do homework! Final Grade of Students Who Took Online Quizzes 2.38 2.49 2.64 2.87 2.98 2.9 3.03 3.04 3.23 3.1 2 2.2 2.4 2.6 2.8 3 3.2 3.4 0 1 2 3 4 5 6 7 8 9 Number of Quizzes Taken Units Used in Calculations Length : A car is 12 feet long, not 12 ! A person is 6 feet tall, not 6 ! Area : A carpet measuring 3 feet(ft) by 4 ft has an area of: ( 3 x 4 )( ft x ft ) = ft2 Speed and Distance : A car traveling 350 miles(mi) in 7 hours(hr) has a speed of: 350 mi / 7 hr = mi / hr In 3 hours the car travels: 3 hr x 50 mi / hr = mi (P. 17) Derived SI Units Quantity Definition of Quantity SI unit Area Length squared m2 Volume Length cubed m3 Density Mass per unit volume kg/m3 Speed Distance traveled per unit time m/s Acceleration Speed changed per unit time m/s2 Force Mass times acceleration of object kg * m/s2 ( =newton, N) Pressure Force per unit area kg/(ms2) ( = pascal, Pa) Energy Force times distance traveled kg * m2/s2 ( = joule, J) How to Solve Chemistry Problems 1) Problem: States all of the information needed to solve the Problem. 2) Plan: Clarify the known and unknown. Suggest the steps needed to find the solution. Develop a roadmap solution. 3)Solution: Calculations appear in the same order as outlined. 4) Check: Is the result what you expect or at least in the same order of magnitude! 5) Comment:Additional information as needed. Conversion Factors : Unity Factors - I Equivalent factors can be turned into conversion factors by dividing one side into the other! 1 mile = 5280 ft or 1 = 1 mile / 5280 ft = 5280 ft / 1 mi 1 in = 2.54 cm or 1 = 1 in / 2.54 cm = 2.54 cm / 1 in In converting one set of units for another, the one desired is on top in the conversion factor, and the old one is canceled out! convert 29,141 ft into miles! 29,141 ft x 1 mi / 5280 ft = mi Conversion Factors - II 1.61 km = 1 mi or 1 = 1.61 km / 1 mi Convert 5.519 miles in to kilometers 5.519 mi x 1.61 km / mi = km conversions in the metric system are easy, as 1 km = 1000 m and 1 meter (m) = 100 centimeters(cm) and 1 cm = 10 millimeters(mm) Therefore: into cm and mm! 8.89 km x 1000m / 1 km = m 8,890 m x 100 cm / m = cm Conversion Factors - III ¥ Multiple conversion factors! ¥ Convert 3.56 lbs/hr into units of ¥ milligrams/sec. 3.56 lbs 1 kg 1000g 1000mg 1 hr 1 min hr 2.205 lbs 1 kg 1 g 60 min 60 sec =x x x x x = mg/s Conversion Factors - IV metric volume to metric volume ¥ 1.35 x 109 km3 = volume of world s oceans ¥ 1.35 x 109 km3 x (103 m/1 km )3 x ( 103 l/m3) = 1.35 x 1021 liters ¥ conversion factors: 1000m = 1km 1000 l = 1m3 Calculate the mass of 1.00 ft3 of Lead (density=11.4 g/ml)? ¥ 1.00 ft3 x (12 in/ft)3 x (2.54 cm/in)3 = ¥ x 11.4 g/cm3 = ¥ Ans. = = Conversion Factors - V Sample Problem 1.5 (p 23) - I Lithium (Li) is a soft, gray solid that has the lowest density of any metal. If a slab of Li weighs 1.49 x 103mg and has sides that measure 20.9 mm by 11.1 mm by 12.0 mm, what is the density of Li in g/ cm3 ? Lengths (mm) of sides Lengths (cm) of sides Mass (mg) of Li Mass (g) of Li Density (g/cm3) of Li Volume (cm3) Sample Problem 1.5 - II Mass (g) of Li = 1.49 x 103 mg x = 1.49 g Length (cm) of one side = 20.9 mm x 1cm / 10 mm = cm Similarly, the other side lengths are 1.11 cm and 1.20 cm Volume (cm3) = cm x 1.11 cm x 1.20 cm = cm3 Density of Li = = 0. g/cm3 cm3 1 g 103 mg 1.49 g Like Sample Problem 1. 5 - Density of a Metal Problem: Cesium is the most reactive metal in the periodic table, what is it s densi ty if a 3.4969 kg cube of Cs has sides of 125.00 mm each? Plan: Calculate the volume from the dimensions of the cube, and calculate the density from the mass and volume. Solution: Volume = (length)3 = (12.500 cm)3 = cm3 length = 125.00 mm = cm density = = = g/ml mass = 3.4969 kg x 1000g/kg = g mass g volume cm3 Temperature Scales and Interconversions Kelvin ( K ) - The Absolut e temperature scale begins at absolute zero and only has positive values. Celsius ( oC ) - The temperature scale used by science, formally called centigrade and most commonly used scale around the world, water freezes at 0oC, and boils at 100oC. Fahrenheit ( oF ) - Commonly used scale in America for our weather reports, water freezes at 32oF, and boils at 212oF. Temperature Conversions T (in K) = T (in 0C) + 273.15 T (in 0C) = T (in K) - 273.15 T (in 0F) = 9/5 T (in 0C) + 32 T (in 0C) = [ T (in 0F) - 32] 5/9 Temperature Conversions - I The boiling point of Liquid Nitrogen is —1950C, what is the temperature in Kelvin and degrees Fahrenheit? T (in K) = T (in 0C) + 273.15 T (in K) = -195.8 0C + 273.15 = K = K T (in 0F) = 9/5 T (in 0C) + 32 T (in 0F) = 9/5 ( -195.8 0C) + 32 = 0F Temperature Conversions - II The normal body temperature is 98.6 0F, what is it in degrees Celsius and Kelvin? T (in 0C) = [ T (in 0F) - 32] 5/9 T (in 0C) = [ 98.6 0F - 32] 5/9 = 0C T (in K) = T (in 0C) + 273.15 T (in K) = 37.0 0C + 273.15 = K Rules for Determining Which Digits Are Significant All digits are significant, except zeros that are used only to position the decimal point. 1. Make sure that the measured quantity has a decimal point. 2. Start at the left of the number and move right until you reach the first nonzero digit. 3. Count that digit and every digit to it s ri ght as significant. Zeros that end a number and lie either after or before the decimal point are significant; thus 1.030 ml has four significant figures, and 5300. L has four significant figures also. Numbers such as 5300 L is assumed to only have 2 significant figures. A terminal decimal point is often used to clearify the situation, but scientific notation is the best! Examples of Significant Digits in Numbers Number - Sig digits Number - Sig digits 0.0050 L two 1.34000 x 107 nm six 18.00 g four 5600 ng two 0.00012 kg two 87,000 L two 83.0001 L six 78,002.3 ng six 0.006002 g four 0.000007800 g four 875,000 oz three 1.089 x 10 -6L four 30,000 kg one 0.0000010048 oz five 5.0000 m3 five 6.67000 kg six 23,001.00 lbs seven 2.70879000 ml nine 0.000108 g three 1.0008000 kg eight 1,470,000 L three 1,000,000,000 g one